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Sarahisme

crazy limit!!

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x->-infinity,lim(x+sqrt(x^2-4x+1))

 

 

i can't for the life of me work out why or how it is possible the limit is 2?!?!

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Edit: Sorry, this was wrong. Thought you said +infinity instead of -infinity :)

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My suggestion in limit solving. Divide the stuff inside the limit by x, you will get: 1 + sqrt(1 - 4/x + 1/x^2). Then substituting in x -> -infinity, you will get 2.

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Huh? If you divide the limit by x, you're also going to need to multiply it by x to keep equality on both sides. That method works well for fractions, but in this case, I think you need something else.

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Huh? If you divide the limit by x, you're also going to need to multiply it by x to keep equality on both sides. That method works well for fractions, but in this case, I think you need something else.

 

 

Aww. you're right XD. I guess I couldn't take the easy way out. Let me think about it ~_~

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x->-infinity' date='lim(x+sqrt(x^2-4x+1))

 

 

i can't for the life of me work out why or how it is possible the limit is 2?!?![/quote']So, we wish to calculate the [math]\lim_{x\to - \infty}{x+\sqrt{x^2-4x+1}[/math][math] = \lim_{x\to\infty}{-x+\sqrt{x^2+4x+1}}[/math].

 

We'll employ a standard trick, multiplication by the 'conjugate'.

 

Note that:

 

[math]-x+\sqrt{x^2+4x+1} = (-x+\sqrt{x^2+4x+1}) \cdot[/math][math]\frac{- x - \sqrt{x^2+4x+1}}{ - x - \sqrt{x^2+4x+1}}=(1)[/math]

 

[math] = \frac{x^2 - (x^2+4x+1)}{-x - \sqrt{x^2+4x+1}} = \frac{-4x-1}{-x - \sqrt{x^2+4x+1}}[/math][math] = \frac{-4x-1}{-x - \sqrt{x^2+4x+1}}[/math][math]\cdot \frac{\frac{1}{x}}{\frac{1}{x}}[/math]

 

[math]=\frac{4+1/x}{1+\frac{1}{x} \sqrt{x^2+4x+1}}[/math][math]=\frac{4+1/x}{1+\sqrt{(\frac{1}{x^2})(x^2+4x+1)}}[/math]

 

[math]=\frac{4+1/x}{1+\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}}[/math]

 

Thus,

 

[math]\lim_{x\to-\infty}{x+\sqrt{x^2-4x+1}=\lim_{x\to\infty}{\frac{4+1/x}{1+\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}}}[/math][math]=\frac{4}{1+\sqrt{1+0+0}}=2[/math]

 

 

Alternatively, we could have used L'Hopital's rule after simplifying step [math](1)[/math], but I wasn't sure if you had learned it as of yet.

 

Edit: On a related note, apparently the [math]\LaTeX[/math] renderer here has a length limit. I had to break up quite a few of the above statements into separate [ math] [/ math] chunks.

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So' date=' we wish to calculate the [math']\lim_{x\to\infty}{x+\sqrt{x^2-4x+1}[/math].

 

We'll employ a standard trick, multiplication by the 'conjugate'.

 

Note that:

 

[math]x+\sqrt{x^2-4x+1} = (x+\sqrt{x^2-4x+1}) \cdot \frac{x - \sqrt{x^2-4x+1}}{x - \sqrt{x^2-4x+1}}(1)[/math]

 

[math] = \frac{x^2 - (x^2-4x+1)}{x - \sqrt{x^2-4x+1}} = \frac{4x-1}{x - \sqrt{x^2-4x+1}} = \frac{4x-1}{x - \sqrt{x^2-4x+1}}[/math][math]\cdot \frac{\frac{1}{x}}{\frac{1}{x}}[/math]

 

[math]=\frac{4-1/x}{1-\frac{1}{x} \sqrt{x^2-4x+1}}[/math][math]=\frac{4-1/x}{1-\sqrt{(\frac{1}{x^2})(x^2-4x+1)}}[/math]

 

[math]=\frac{4-1/x}{1-\sqrt{1-\frac{4}{x}+\frac{1}{x^2}}}[/math]

 

Thus,

 

[math]\lim_{x\to\infty}{x+\sqrt{x^2-4x+1}=\lim_{x\to\infty}{\frac{4-1/x}{1-\sqrt{1-\frac{4}{x}+\frac{1}{x^2}}}}[/math][math]=\frac{4}{1-\sqrt{1-0+0}}=2[/math]

 

 

Alternatively, we could have used L'Hopital's rule after simplifying step [math](1)[/math], but I wasn't sure if you had learned it as of yet.

 

Edit: On a related note, apparently the [math]\LaTeX[/math] renderer here has a length limit. I had to break up quite a few of the above statements into separate [ math] [/ math] chunks.

 

Dapthar this utterly confused me. Firstly the limit is towards negative infinity, not positive infinity. Second of all, in the denominator you have 1-1=0, which is division by zero error. Not two in the denominator.

 

You seem to have written her problem down incorrectly. Let me give it a try:

 

[math] \lim_{x\to - \infty} [x + \sqrt{x^2-4x+1}] [/math]

 

So you want to know what happens to the value of the function you are taking the limit of, as x becomes increasingly negative.

 

The first thing I would do, is replace x by -x, and take the limit as x goes to positive infinity. So you have this instead...

 

[math] \lim_{x\to \infty} [-x + \sqrt{x^2+4x+1}] [/math]

 

Intuitively, x^2 dominates over 4x. In other words, as x becomes increasingly huge, x^2 becomes enormous in comparison to 4x, so that we can neglect the addition of 4x to x^2, and certainly adding 1 doesn't alter the answer much either. Thus, we should expect the limit above to be equal to the following limit:

 

[math] \lim_{x\to \infty} [-x + \sqrt{x^2}] = \lim_{x\to \infty} [-x + x] = \lim_{x\to \infty} [0] = 0 [/math]

 

Now, of course this is not rigorous, but it's intuitive.

 

And one other thing, the square root of x^2 is the absolute value of x.

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Dapthar this utterly confused me. Firstly the limit is towards negative infinity' date=' not positive infinity. Second of all, in the denominator you have 1-1=0, which is division by zero error. Not two in the denominator.

[/quote']You're right Johnny, I'll go back and fix my post.

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You're right Johnny, I'll go back and fix my post.

No problem Dapthar, but...

 

Isn't the limit zero?

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o.O

 

The problem with the negative infinity can be easily resolved with the t = -1/x substitution, where as x -> negative infinity, t -> 0. If you do this substitution on his third line will give you good results.

 

An alternative method I thought of was to actually factor the insides of the square root before doing all of that, you will find that it turns out to be something like: (x-a)(x+a), where a = 2 (+-) sqrt3. Then from there doing the same, you'll end up with a cleaner process.

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o.O

 

The problem with the negative infinity can be easily resolved with the t = -1/x substitution' date=' where as x -> negative infinity, t -> 0. If you do this substitution on his third line will give you good results.

 

An alternative method I thought of was to actually factor the insides of the square root before doing all of that, you will find that it turns out to be something like: (x-a)(x+a), where a = 2 (+-) sqrt3. Then from there doing the same, you'll end up with a cleaner process.[/quote']

 

Yes I thought about that too.

 

(x+a)(x+b) = x^2 +xb+ax+ab

 

ab=1

 

x(a+b)=4x

(a+b) =4

 

b=1/a

 

(a+1/a)=4

 

(a^2+1)/a = 4

 

a^2+1=4a

a^2-4a+1=0

 

Now use quadratic formula. By a theorem due to Abel, there are exactly two roots of the polynomial above. The roots of the polynomial above are given by:

 

[math] \frac{4 + \sqrt{16-4}}{2} [/math]

 

[math] \frac{4 - \sqrt{16-4}}{2} [/math]

 

Now

 

[math] \sqrt{16-4} = \sqrt{12} = \sqrt{3*2*2} = 2 \sqrt{3} [/math]

 

Thus, we can write the two roots in a simpler form:

 

[math] \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}[/math]

 

[math] \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} [/math]

 

Let us write this:

 

[math] r1 = 2 + \sqrt{3}[/math]

 

[math] r2 = 2 - \sqrt{3} [/math]

 

So if a=r1 then the polynomial statement is true, and if a=r2, then the polynomial statement is true, and there are no other numbers which when instantiated for the variable a... yield a true statement.

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Okay, here's my kind of "sketch" proof:

 

[math]x^2 - 4x + 1 = (x-2)^2 - 3[/math]

 

Now let [math]y = x-2[/math], so the limit becomes:

 

[math]\lim_{y\to -\infty} \left( 2 + y + \sqrt{y^2 -3} \right)[/math]

 

Now, [math]2 + y + \sqrt{y^2 -3} = 2 + \frac{3}{y - \sqrt{y^2 - 3}}[/math]

 

The fraction on the right tends to zero, so the limit is 2. I did it Dapthar's way at first, but got unstuck at the end also. What do you guys think? Personally, I think it's a little bit dodgy and I haven't actually proved that the fraction on the right tends to zero, but I think that it's pretty easy to do so (sandwich rule?)

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Edit: On a related note, apparently the [math]\LaTeX[/math] renderer here has a length limit. I had to break up quite a few of the above statements into separate [ math] [/ math][/b'] chunks.

 

Indeed. The limit is about 500 pixels to stop the page breaking up and whatnot.

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Ok, I fixed my earlier calculation. (See above post.[/i'])

 

There are still a few errors Dapthar. Look at the sign of your 4x term. In one of the steps you switch from negative to positive by accident.

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There are still a few errors Dapthar. Look at the sign of your 4x term. In one of the steps you switch from negative to positive by accident.
I cancel out the [math]-1[/math] from the numerator and the denominator in this step:

 

[math]\frac{-4x-1}{-x - \sqrt{x^2+4x+1}}[/math][math]\cdot \frac{\frac{1}{x}}{\frac{1}{x}}[/math]

 

[math]=\frac{4+1/x}{1+\frac{1}{x} \sqrt{x^2+4x+1}}[/math]

 

If that's the error you're referring to.

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I cancel out the [math]-1[/math] from the numerator and the denominator in this step:

 

[math]\frac{-4x-1}{-x - \sqrt{x^2+4x+1}}[/math][math]\cdot \frac{\frac{1}{x}}{\frac{1}{x}}[/math]

 

[math]=\frac{4+1/x}{1+\frac{1}{x} \sqrt{x^2+4x+1}}[/math]

 

If that's the error you're referring to.

 

 

That's not an error. Here i will show you what I'm referring to...

[math] \frac{-4x-1}{-x - \sqrt{x^2-4x+1}}[/math][math] = \frac{-4x-1}{-x - \sqrt{x^2+4x+1}}[/math][math]\cdot \frac{\frac{1}{x}}{\frac{1}{x}}[/math]

 

You can see this is false rather rapidly.

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Okay this is based on Dapthar's original post. I orignally had a slightly different method in mind, but I also started out using the multiplication by the conjugate.

 

[math]\lim_{x\to\infty}{x+\sqrt{x^2-4x+1}[/math].

 

[math]x+\sqrt{x^2-4x+1} = (x+\sqrt{x^2-4x+1}) \cdot \frac{x - \sqrt{x^2-4x+1}}{x - \sqrt{x^2-4x+1}}(1)[/math]

 

[math] = \frac{x^2 - (x^2-4x+1)}{x - \sqrt{x^2-4x+1}} = \frac{4x-1}{x - \sqrt{x^2-4x+1}} = \frac{4x-1}{x - \sqrt{x^2-4x+1}}[/math][math]\cdot \frac{\frac{1}{x}}{\frac{1}{x}}[/math]

 

[math]=\frac{4+1/x}{1+\frac{1}{x} \sqrt{x^2+4x+1}}[/math]

 

let t = -1/x, such that [math] \lim_{x\to-\infty} = \lim_{t\to0} [/math]

 

[math]=\frac{4+t}{1+t \sqrt{\frac{1}{t^2}+\frac{4}{t}+1}}[/math]

[math]=\frac{4+t}{1+\sqrt{1+4t+t^2}}[/math]

 

By substituting t = 0, we get our desired 2.

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From what I've seen, the methods are pretty much all the same. Dapthar's is probably the best because it doesn't involve any substitution.

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That's not an error. Here i will show you what I'm referring to...

 

 

You can see this is false rather rapidly.

Yup, you're right. When I was going back through and changing all the signs I missed a few. I went back and edited my previous post, and fixed the signs again.

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From what I've seen, the methods are pretty much all the same. Dapthar's is probably the best because it doesn't involve any substitution.

 

Actually, I just followed your proof, and it convinced me that the limit is 2. A good proof must be:

 

1. Utterly convincing

2. Short

 

I still have to look at Dapthar's. But I mean I remember that technique, multiplying by the conjugate, because the interior term vanishes.

 

Ok, I just followed Dapthar's as well, and both were convincing. I would say this... some folks will follow one better than the other, so I would do both. :)

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Indeed. Always go for the way that you understand best. Comes out trumps every time (unless it's wrong :P).

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Indeed. Always go for the way that you understand best. Comes out trumps every time (unless it's wrong :P).

 

Indeed. :)

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[math] \lim_{x\to \infty} [-x + \sqrt{x^2+4x+1}] [/math]

 

Intuitively' date=' x^2 dominates over 4x. In other words, as x becomes increasingly huge, x^2 becomes enormous in comparison to 4x, so that we can neglect the addition of 4x to x^2, and certainly adding 1 doesn't alter the answer much either. Thus, we should expect the limit above to be equal to the following limit:

 

[math'] \lim_{x\to \infty} [-x + \sqrt{x^2}] = \lim_{x\to \infty} [-x + x] = \lim_{x\to \infty} [0] = 0 [/math]

 

Now, of course this is not rigorous, but it's intuitive.

 

And one other thing, the square root of x^2 is the absolute value of x.

I haven't seen anyone explain why this is wrong yet, even though everyone seems to agree that the limit is 2. :)

 

What you're doing is factoring out an x from the square root, so you get this:

 

[math] \lim_{x\to -\infty} [x + |x| \sqrt{1+\frac{4}{x}+\frac{1}{x^2}}] [/math]

 

So far so good. But then you're saying that what's left under the square root sign is going towards one as x goes to negative infinity, so that the limit above should go towards [math] \lim_{x\to -\infty} [x - x] = 0 [/math]. This would be true if x were finite -- but since x in this case is approaching infinity, multiplying it with something that's just approaching one doesn't necessarily leave it unchanged. It could converge to anything, or diverge, for all we know.

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